Answer:
The p-value for this test is 0.22065.
Step-by-step explanation:
We are given that a national study report indicated that 20.9% of Americans were identified as having medical bill financial issues.
A news organization randomly sampled 400 Americans from 10 cities and found that 90 reported having such difficulty.
<u><em>Let p = proportion of Americans who were identified as having medical bill financial issues in 10 cities.</em></u>
SO, Null Hypothesis,
: p
20.9% {means that % of Americans who were identified as having medical bill financial issues in these 10 cities is less than or equal to 20.9%}
Alternate Hypothesis,
: p > 20.9% {means that % of Americans who were identified as having medical bill financial issues in these 10 cities is more than 20.9% and is more severe}
The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;
T.S. =
~ N(0,1)
where,
= sample proportion of 400 Americans from 10 cities who were found having such difficulty =
= 0.225 or 22.5%
n = sample of Americans = 400
So, <u><em>test statistics</em></u> = ![\frac{0.225-0.209}{{\sqrt{\frac{0.225(1-0.225)}{400} } } } }](https://tex.z-dn.net/?f=%5Cfrac%7B0.225-0.209%7D%7B%7B%5Csqrt%7B%5Cfrac%7B0.225%281-0.225%29%7D%7B400%7D%20%7D%20%7D%20%7D%20%7D)
= 0.77
<u></u>
<u>Now, P-value of the test statistics is given by the following formula;</u>
P-value = P(Z > 0.77) = 1 - P(Z
0.77)
= 1 - 0.77935 = 0.22065